A limit

[Tolaso J Kos]

Prove that:

$$\lim_{n\rightarrow +\infty}\sqrt[n^2]{\binom{n+1}{1}\binom{n+2}{2}\cdots\binom{n+n}{n}}=\frac{4}{\sqrt{e}}$$

[akotronis]

Let's take it a little bit further:

If \(A_n=\sqrt[n^2]{\binom{n+1}{1}\binom{n+2}{2}\cdots\binom{n+n}{n}}\), show that

\[\displaystyle{A_n=\frac{4}{\sqrt{e}}\left(1-\frac{\ln n}{2n}+\frac{\ln(4/\sqrt{2\pi})}{n}+\frac{\ln^2n}{8n^2}+\mathcal O\left(\frac{\ln n}{n^2}\right)\right)}\]

[r9m]

\begin{align*}&\frac{1}{n^2}\log \left(\prod\limits_{j=1}^{n}\binom{n+j}{n}\right) \\&= \frac{1}{n^2}\log \left(\prod\limits_{j=1}^{n}\frac{(n+j)_{n}}{j!}\right)\\&= \frac{1}{n^2}\sum\limits_{j=1}^{n}(n+1-j)\log(n+j) - \frac{1}{n^2}\sum\limits_{j=1}^{n}j\log (n+1-j)\\&= \frac{1}{n}\sum\limits_{j=1}^{n} \left(1-\frac{j-1}{n}\right)\log \left(1+\frac{j}{n}\right) - \frac{j}{n}\log \left(1-\frac{j-1}{n}\right)\end{align*}

Thus, replacing the limit of Riemann sum with integrals: $$\lim\limits_{n \to \infty} \frac{1}{n^2}\log \left(\prod\limits_{j=1}^{n}\binom{n+j}{n}\right) = \int_0^1 (1-x)\log(1+x) - x\log(1-x)\,dx = 2\log 2 - \frac{1}{2}$$

Hence, $$\lim_{n\rightarrow +\infty}\sqrt[n^2]{\binom{n+1}{1}\binom{n+2}{2}\cdots\binom{n+n}{n}}=\frac{4}{\sqrt{e}}$$

As for akotronis's strong asymptotics I guess we could start with Euler-Maclaurin summation formula.